image.png
0. 链接
1. 题目
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Example 1:
Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Not 7-1 = 6, as selling price needs to be larger than buying price.
Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
2. 思路1: 动态规划
记录max_profit, buy_price
从左到右遍历, 遇到price比buy_price小的, 则更新buy_price=price; 遇到大的，则试图更新max_profit=max(max_profit, price-buy_price)
时间复杂度: ```O(N)``
空间复杂度:
O(1)
3. 代码
# coding:utf8
from typing import List
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if len(prices) == 0:
return 0
max_profit = 0
buy_price = prices[0]
for price in prices:
if price < buy_price:
buy_price = price
if price > buy_price:
max_profit = max(max_profit, price - buy_price)
return max_profit
def my_test(solution, prices):
print('input: {}; output: {}'.format(prices, solution.maxProfit(prices)))
solution = Solution()
my_test(solution, [7, 1, 5, 3, 6, 4])
my_test(solution, [7, 6, 4, 3, 1])
my_test(solution, [2, 4, 1])
输出结果
input: [7, 1, 5, 3, 6, 4]; output: 5
input: [7, 6, 4, 3, 1]; output: 0
input: [2, 4, 1]; output: 2
4. 结果
image.png
